根据前，中（后，中）构建二叉树

1，力扣105

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int p_len = preorder.length;
        int i_len = inorder.length;
        if(p_len==0||i_len==0){
            return null;
        }
        TreeNode root = new TreeNode();
        root.val = preorder[0];
        int k = 0;
        for(int i = 0;i<i_len;i++){
            if(inorder[i]==root.val)
           { k = i;
            break;
           }//记得加括号
        }
        //有点繁琐但易于理解
        int[] left_p = Arrays.copyOfRange(preorder,1,k+1);
        int[] left_i = Arrays.copyOfRange(inorder,0,k);
        root.left = buildTree(left_p,left_i);//构建左边的子树
        int[] right_p = Arrays.copyOfRange(preorder,k+1,p_len);
        int[] right_i = Arrays.copyOfRange(inorder,k+1,i_len);
        root.right = buildTree(right_p,right_i);//构建右子树 
        return root;
    }
}

2，力扣106

给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        int i_len = inorder.length;//获取数组属性，注意length与length()的区别
        int p_len = postorder.length;
        if(i_len==0||p_len==0)
        return null;
        //找根节点
        int rootValue = postorder[p_len-1];
        TreeNode root = new TreeNode(rootValue);
        //找中间节点
        int k = 0;
        for(int i=0;i<i_len;i++){
            if(inorder[i]==rootValue){
                k=i;
                break;
            }
        }
        //
        int[] left_i = Arrays.copyOfRange(inorder,0,k);
        int[] left_p = Arrays.copyOfRange(postorder,0,k);
        root.left = buildTree(left_i,left_p);
        int[] right_i = Arrays.copyOfRange(inorder,k+1,i_len);
        int[] right_p = Arrays.copyOfRange(postorder,k,p_len-1);//前k个元素以及被取走，剩下的从第k个开始就是右子树的节点了
        root.right = buildTree(right_i,right_p);
        return root;

    }
}